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How can we compute the Fourier transform of the function: $$ f(x) = \frac{1}{(1+x^2)^2}. $$ Is the resulting function differentiable? If so, how many times?

I'm thinking if this would require partial fractions?

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up vote 2 down vote accepted

We are trying to compute $$ \begin{align} \int_{-\infty}^\infty\frac{e^{-2\pi ixt}}{(1+x^2)^2}\mathrm{d}x &=\int_{-\infty}^\infty\frac{e^{2\pi ixt}}{(1+x^2)^2}\mathrm{d}x\\ &=\int_{-\infty}^\infty\frac{t^3e^{2\pi ix}}{(t^2+x^2)^2}\mathrm{d}x\\ &=\int_\gamma\frac{t^3e^{2\pi iz}}{(t^2+z^2)^2}\mathrm{d}z\tag{1} \end{align} $$ where $\gamma$ is the contour that goes from $-L$ to $L$ along $\mathbb{R}$ then circles back counterclockwise to $-L$ through the upper half plane. Note that the size of the integrand on the circular part of the contour is $\le t^3/L^4$ and the length of the length of the circular part is $\pi L$, thus the integral over the contour goes to $0$ as $L\to\infty$. Also note that the integral does not depend on the sign of $t$, so we can assume $t>0$ and replace $t$ by $|t|$ in the final result.

By partial fractions, we get $$ \begin{align} \frac{t^3}{(t^2+z^2)^2} &=\frac{t^3}{(z+it)^2(z-it)^2}\\ &=\frac{-t/4}{(z-it)^2}+\frac{-t/4}{(z+it)^2}+\frac{t/2}{(z-it)(z+it)}\\ &=\frac{-t/4}{(z-it)^2}+\frac{-t/4}{(z+it)^2}-\frac{i/4}{z-it}+\frac{i/4}{z+it}\tag{2} \end{align} $$ Puting $(1)$ and $(2)$ together yields $$ \begin{align} \int_{-\infty}^\infty\frac{e^{2\pi ixt}}{(1+x^2)^2}\mathrm{d}x &=\int_\gamma\frac{-t/4\;e^{2\pi iz}}{(z-it)^2}\mathrm{d}z +\int_\gamma\frac{-t/4\;e^{2\pi iz}}{(z+it)^2}\mathrm{d}z\\ &+\int_\gamma\frac{-i/4\;e^{2\pi iz}}{z-it}\mathrm{d}z +\int_\gamma\frac{i/4\;e^{2\pi iz}}{z+it}\mathrm{d}z\\[6pt] &=2\pi i(-t/4)(2\pi i)e^{-2\pi t}+0\\[6pt] &+2\pi i(-i/4)e^{-2\pi t}+0\\[6pt] &=\pi^2t\;e^{-2\pi t}+\pi/2\;e^{-2\pi t}\tag{3} \end{align} $$ Thus, the final answer is $$ \int_{-\infty}^\infty\frac{e^{-2\pi ixt}}{(1+x^2)^2}\mathrm{d}x =\frac\pi2(1+2\pi|t|)\;e^{-2\pi|t|}\tag{4} $$ Analysis of $\hat{f}$

$(\pi/2+\pi^2t)\;e^{-2\pi t}$ is $C^\infty(\mathbb{R})$. However, due to the discontinuity in the derivative of $|t|$ at $t=0$, care must be taken with the derivatives of $\hat{f}(t)=(\pi/2+\pi^2|t|)\;e^{-2\pi|t|}$ at $t=0$.

Taking the first and second derivatives of the formula in $(4)$ yields $$ \frac{\mathrm{d}}{\mathrm{d}t}\hat{f}(t)=-2\pi^3t\,e^{-2\pi|t|}\to0\quad\text{as}\quad t\to0\tag{5} $$ and $$ \frac{\mathrm{d}^2}{\mathrm{d}t^2}\hat{f}(t)=-2\pi^3(1-2 \pi|t|)\,e^{-2 \pi|t|}\to-2\pi^3\quad\text{as}\quad t\to0\tag{6} $$ However, the third derivative is discontinuous at $t=0$. $$ \frac{\mathrm{d}^3}{\mathrm{d}t^3}\hat{f}(t)=8\pi^4(\rm{sgn}(t)-\pi t)\,e^{-2 \pi|t|}\tag{7} $$ Another way to see that the first and second derivatives exist is that $$ \frac{\mathrm{d}}{\mathrm{d}t}\int_{-\infty}^\infty\frac{e^{-2\pi ixt}}{(1+x^2)^2}\mathrm{d}x =\int_{-\infty}^\infty\frac{-2\pi ix\,e^{-2\pi ixt}}{(1+x^2)^2}\mathrm{d}x\tag{8} $$ and $$ \frac{\mathrm{d}^2}{\mathrm{d}t^2}\int_{-\infty}^\infty\frac{e^{-2\pi ixt}}{(1+x^2)^2}\mathrm{d}x =\int_{-\infty}^\infty\frac{-4\pi^2x^2\,e^{-2\pi ixt}}{(1+x^2)^2}\mathrm{d}x\tag{9} $$ converge absolutely.

However, the third derivative is not guaranteed to exist since $$ \frac{\mathrm{d}^3}{\mathrm{d}t^3}\int_{-\infty}^\infty\frac{e^{-2\pi ixt}}{(1+x^2)^2}\mathrm{d}x =\int_{-\infty}^\infty\frac{8\pi^3ix^3\,e^{-2\pi ixt}}{(1+x^2)^2}\mathrm{d}x\tag{10} $$ does not converge absolutely.

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I assume you know that the Fourier transform of $1/(1+x^2)$ is a constant multiple of $g(\xi)=e^{-|\xi|}$ (the exact form may depend on the definition of Fourier transform you are using). By the convolution theorem, $\hat f$ will be a constant multiple of $$ g\ast g(\xi)=(1+|\xi|)e^{-|\xi|}. $$ $\hat f$ is of class $C^2$ but not $C^3$. You do not need to compute $\hat f$ to know this. It is enough to realize that $|x|^2f\in L^1(\mathbb{R})$ but $|x|^3f\not\in L^1(\mathbb{R})$.

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By residue theory, $$\forall t\in\mathbb{R}^+,\quad \int_{-\infty}^{+\infty}\frac{\cos(tx)}{(1+x^2)^2}\,dx = \frac{\pi}{2}\,(t+1)\,e^{-t},$$ so: $$\int_{-\infty}^{+\infty}\frac{\cos(tx)}{(1+x^2)^2}\,dx = \frac{\pi}{2}\,(|t|+1)\,e^{-|t|}.$$

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